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SSSSS [86.1K]
2 years ago
15

HELP ME PLEASE!!

Mathematics
1 answer:
DiKsa [7]2 years ago
5 0
C=A+B
C₄₁= row=4;  column=1
C₄₁=A₄₁+B₄₁         
=2(7)-3(4)
=14-12
=2

Answer: C₄₁=2
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Please help<br><br> x-3=x^(2)-10x+25
schepotkina [342]

Answer:

x=7,4

Step-by-step explanation:

8 0
2 years ago
Help!!!!!!!!!!<br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B500%7D%20%7D%7B%5Csqrt%5B%5D%7B5%7D%20%7D" id="TexFormu
DochEvi [55]

Answer:

10

Step-by-step explanation:

Rationalize the denominator by multiplying both the top and bottom by sqrt(5). you have sqrt(500) * sqrt(5) / sqrt(5)^2

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7 0
2 years ago
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What percent of a data set is represented by Q1
Ludmilka [50]
I think it's 25%

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4 0
3 years ago
Read 2 more answers
A car travels 3 times around a traffic circle whose radius is 80 feet. What is the distance the car will travel? Use 3.14 for π
Harman [31]

Answer : Distance will be 1507.2 feet .

Explanation :

Since we have given that

Radius of circle = 80 feet

So,

Circumference of circle is given by

2\pi r=2\times 3.14\times 80=502.4 \text{ feet}

Since , a car travels 3 times around a traffic circle.

So,

\text{ Distance covered by the car will travel}= 3\times 502.4= 1507.2 \text{ feet }

So, Distance will be 1507.2 feet .

5 0
3 years ago
If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi
Finger [1]

<u>ANSWER: </u>

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

<u>SOLUTION: </u>

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = \mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2} --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval  

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16

34 \times 1+\frac{1}{2} \times a \times 1^{2}=18

34+\frac{a}{2}=18

\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0
3 years ago
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